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Question: A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A sma...

A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ\theta with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is μ\mu. Find the range of the angular speed for which the block will not slip.

Answer

Range of angular speed is [gRsinθtanθμ1+μtanθ,gRsinθtanθ+μ1μtanθ]\left[ \sqrt{\frac{g}{R\sin\theta} \frac{\tan\theta - \mu}{1 + \mu\tan\theta}}, \sqrt{\frac{g}{R\sin\theta} \frac{\tan\theta + \mu}{1 - \mu\tan\theta}} \right] with appropriate conditions for μ\mu and θ\theta.

Explanation

Solution

The problem involves finding the range of angular speeds for which a block in a rotating hemispherical bowl does not slip. This is determined by analyzing the forces acting on the block at the minimum and maximum angular speeds, considering static friction.

Case 1: Minimum Angular Speed (ωmin\omega_{min})

At this speed, the block tends to slip downwards, so friction acts upwards. We resolve forces into horizontal and vertical components to find the normal force NN and then ωmin\omega_{min}.

Vertical Equilibrium: Ncosθ+fssinθmg=0N\cos\theta + f_s\sin\theta - mg = 0

Horizontal Motion (Centripetal Force): Nsinθfscosθ=mωmin2RsinθN\sin\theta - f_s\cos\theta = m\omega_{min}^2 R\sin\theta

Solving these equations, we find: ωmin=gRsinθtanθμ1+μtanθ\omega_{min} = \sqrt{\frac{g}{R\sin\theta} \frac{\tan\theta - \mu}{1 + \mu\tan\theta}} (for tanθμ\tan\theta \ge \mu)

If tanθ<μ\tan\theta < \mu, ωmin=0\omega_{min} = 0.

Case 2: Maximum Angular Speed (ωmax\omega_{max})

At this speed, the block tends to slip upwards, so friction acts downwards. We again resolve forces into horizontal and vertical components.

Vertical Equilibrium: Ncosθfssinθmg=0N\cos\theta - f_s\sin\theta - mg = 0

Horizontal Motion (Centripetal Force): Nsinθ+fscosθ=mωmax2RsinθN\sin\theta + f_s\cos\theta = m\omega_{max}^2 R\sin\theta

Solving these equations, we find: ωmax=gRsinθtanθ+μ1μtanθ\omega_{max} = \sqrt{\frac{g}{R\sin\theta} \frac{\tan\theta + \mu}{1 - \mu\tan\theta}} (for tanθ<1/μ\tan\theta < 1/\mu)

If tanθ1/μ\tan\theta \ge 1/\mu, ωmax=\omega_{max} = \infty.

Therefore, the range of angular speed is [ωmin,ωmax][\omega_{min}, \omega_{max}].